Question

我正在运行此查询：

SELECT DISTINCT "items"."id" 
FROM "items" 
LEFT OUTER JOIN "item_explicit_mods" ON "item_explicit_mods"."item_id" = "items"."id" 
LEFT OUTER JOIN "explicit_mods" ON "explicit_mods"."id" = "item_explicit_mods"."explicit_mod_id" 
LEFT OUTER JOIN "item_implicit_mods" ON "item_implicit_mods"."item_id" = "items"."id" 
LEFT OUTER JOIN "implicit_mods" ON "implicit_mods"."id" = "item_implicit_mods"."implicit_mod_id" 
LEFT OUTER JOIN "shops" ON "shops"."id" = "items"."shop_id" 
WHERE ((item_explicit_mods.explicit_mod_id = 35 
        AND item_explicit_mods.primary_value >= 5 AND item_explicit_mods.primary_value <= 6) 
    OR (item_explicit_mods.explicit_mod_id = 48)) 
GROUP BY items.id 
HAVING COUNT(item_explicit_mods.id) = 2 
ORDER BY "items"."created_at" 
ASC LIMIT 100

Sqlite解释正在产生这个

0|0|0|SCAN TABLE items USING INTEGER PRIMARY KEY (~1000000 rows)
0|1|1|SEARCH TABLE item_explicit_mods USING AUTOMATIC COVERING INDEX (explicit_mod_id=?) (~7 rows)
0|1|1|SEARCH TABLE item_explicit_mods USING AUTOMATIC COVERING INDEX (explicit_mod_id=?) (~7 rows)
0|2|2|SEARCH TABLE explicit_mods USING INTEGER PRIMARY KEY (rowid=?) (~1 rows)
0|3|3|SEARCH TABLE item_implicit_mods USING AUTOMATIC COVERING INDEX (item_id=?) (~7 rows)
0|4|4|SEARCH TABLE implicit_mods USING INTEGER PRIMARY KEY (rowid=?) (~1 rows)
0|5|5|SEARCH TABLE shops USING INTEGER PRIMARY KEY (rowid=?) (~1 rows)
0|0|0|USE TEMP B-TREE FOR DISTINCT
0|0|0|USE TEMP B-TREE FOR ORDER BY

此查询正在执行＆gt;运行10秒。知道为什么它在主键上做SCAN吗？

Answer 1

知道它为什么在主键上进行扫描？

我想因为你这样做了：

SELECT DISTINCT "items"."id"

该表的列id是主键。您正在寻找引用连接中其他表的不同标识符。因此，主键上有一个SCAN。

此外，在您只是寻找不同的标识符时，为什么要将结果分组到GROUP BY items.id？

Answer 2

尝试将您的查询更改为此

SELECT "items"."id" 
FROM "items" 
INNER JOIN "item_explicit_mods" ON "item_explicit_mods"."item_id" = "items"."id" AND (((item_explicit_mods.explicit_mod_id = 35 
        AND item_explicit_mods.primary_value >= 5 AND item_explicit_mods.primary_value <= 6) 
    OR (item_explicit_mods.explicit_mod_id = 48)) )
LEFT OUTER JOIN "explicit_mods" ON "explicit_mods"."id" = "item_explicit_mods"."explicit_mod_id" 
LEFT OUTER JOIN "item_implicit_mods" ON "item_implicit_mods"."item_id" = "items"."id" 
LEFT OUTER JOIN "implicit_mods" ON "implicit_mods"."id" = "item_implicit_mods"."implicit_mod_id" 
LEFT OUTER JOIN "shops" ON "shops"."id" = "items"."shop_id"  
GROUP BY items.id 
HAVING COUNT(item_explicit_mods.id) = 2 
ORDER BY "items"."created_at" 
ASC LIMIT 100

为什么这个SQL查询太慢了？

2 个答案: